# [Haskell-beginners] list splitting - nice implementation?

Emmanuel Touzery etouzery at gmail.com
Sun Nov 18 11:06:57 CET 2012

```That is EXACTLY the kind of answer i was hoping for!
Great implementation, I'll be sure to reuse that trick of zipping with the
tail of the list.. really really good.

I'm relieved it's not trivial (for me) to write since i could not come up
with it, and happy i understand it :-)

Intuitively it's more expensive than what an imperative language would do
(new list creations, going several times over the list -zip, spam, map -
still O(n) though).

If it was in your project you'd use that, or would you use a more
"straightforward" implementation and you came up with this one just because
i asked explicitely for such a way?
On 18 Nov 2012 10:47, "Tobias Brandt" <tob.brandt at googlemail.com> wrote:

> Here is a possible solution:
>
> import Data.List
>
> split xs = getSnds \$ span (uncurry (<)) \$ zip xs (tail xs)
>     where getSnds (as, bs) = (map snd as, map snd bs)
>
> firstPart xs = fst \$ split xs
>
> sndPart xs = snd \$ split xs
>
> This is a one pass algortihm, it works for infinite lists.
>
> On 18 November 2012 08:45, Emmanuel Touzery <etouzery at gmail.com> wrote:
>
>> Hello,
>>
>>  i wonder what would be the idiomatic way to achieve that algorithm in
>>
>> [1,4,56,450,23,46,52] => [1,4,56,450]
>> [1,4,56,450,23,46,52] => [23,46,52]
>>
>>  in other words split the list when one element gets smaller than the
>> previous one. Tge rest of the time the list is sorted. There would be only
>> two lists, not N. I always need the first or second sublist, I don't need
>> both at once. But of course a more complete algorithm handling the N case
>> and/or returning both sublists would be good.
>>
>>  i could code this by hand, but i'm trying to use as much as possible
>> builtin higher-order functions. However in this case so far I've only come
>> up with this:
>>
>> import Data.List
>>
>> isSorted :: Ord a => [a] -> Bool
>> isSorted l = (sort l) == l
>>
>> secondPart :: Ord a => [a] -> [a]
>> secondPart l = head \$ filter isSorted (tails l)
>>
>> firstPart :: Ord a => [a] -> [a]
>> firstPart l = last \$ filter isSorted (inits l)
>>
>>  It is concise alright, but it seems contrived and also in terms of
>> performance I don't think it's OK (for small lists sure but for big lists?).
>>
>>  Anyway, somehow I think something as simple as this must be doable very
>> concisely and with optimal performance using only builtin higher-order
>> functions. Any idea?
>>
>>  Thanks!
>>
>> Emmanuel
>>
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